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Given the potential barrier, \begin{align} V(x, y) = \left\{ \begin{array}{cc} V_{0} & \hspace{5mm} \text{if $0 \leq x \leq D$} \\ 0 & \hspace{5mm} \text{otherwise} \end{array} \right. \end{align}

the Hamiltonian of the system is $$\hat H = -\frac{\hbar^{2}}{2m}\nabla^{2}+V$$

Hence for $x<0$, the time-independent wavefunction is: $$\Psi(x) = A\,exp(ikx)+B\, exp(-ikx)$$

This is an eigenvector of $\hat H$ with the first term representing the incident wave while the second term represents the reflected wave.

Now for this region $[\hat H, \hat p]=0$, so they should have common non-degenerate eigenvectors.

But the above wavefunction is not an eigenvector of $\hat p$. What am I thinking wrong here?

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    $\begingroup$ You might want to check this out : physics.stackexchange.com/q/221007. That the commutator is zero does not mean that both operators have the same eigenvectors. What they do posses is a set of simultaneous eigenvectors. $\endgroup$ – WarreG Jun 12 at 13:32
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In your example the difficulty is that you're taking linear combination of eigenstates of $p$ but with different eigenvalues, so the resulting combination is no longer an eigenstate of $p$, even if the pieces are separately eigenstates.

An alternate example would be the simple case $[\hat H,\hat L^2]=0$ and a hydrogen atom state with $n=2$ so that $\ell=0,1$ can occur. Then $\{\vert n\ell m\rangle\}$ are simultaneous eigenvectors of $\hat H$ and $\hat L^2$ but a combination of these containing different $\ell$s will not be a simultaneous eigenstate of both since different $\ell$ states have different eigenvalues of $\hat L^2$.

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  • $\begingroup$ Great example and great answer. $\endgroup$ – gented Jun 14 at 14:17
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If the commutator $[H,p] = 0$, it means that there exists a system of eigenvectors which are common to both operators, and that system of eigenvectors spans the Hilbert space of both the operators.

Here the system of simultaneous eigenvectors of $H$ and $p$ are $\{e^{ikx}\}$ for all $k \in (-\infty, \infty)$. We call this continuous eigenspectrum. Now if you look at your solution wavefunction, you will see that the solution is a linear superposition of two eigenvectors. This superposition state may not be an eigenvector for the momentum space.

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$\def\hp{{\hat p}} \def\hH{{\hat H}}$ IMO there is a basic misunderstanding in your arguments, about what operators and eigenfunctions do mean.

You write

Hence for $x<0$, the time-independent wavefunction is: $\Psi(x)=A\,\exp(ikx)+B\,\exp(-ikx)$

This is an eigenvector of $\hat H$

No, this is at least an improper way of saying. It's true that Schr. eqn for stationary states requires a solution of that form for $x<0$. But you can't call it an eigenvector. This name is properly used only for a function obeying Schr. eqn on the whole real line.

Analogously, you're not allowed to say

for this region $[\hH,\hp]=0$

Operators are defined over a space of functions and no significant conclusion can be drawn from a relationship only holding over a subset of functions domain.

so they should have common non-degenerate eigenvectors

This conclusion would be right if your operators did really commute, all over Hilbert space, not on functions restricted to a subset of real line.

Note that if $\hH$ and $\hp$ did commute they would have a complete system of common eigenfunctions. This happens for a free particle: here eigenfunctions of $\hp$ are $\exp(ikx)$ for $k\in\Bbb R$. This is a complete set and all are also eigenfunctions of $\hH$ although degenerate on $k$ sign.

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  • $\begingroup$ You're technically correct but your answer would be much improved if you could explain domain issues in connection to the whole real line. $\endgroup$ – ZeroTheHero Jun 13 at 15:15
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The wave-function you have written down is the solution to the "free" wave equation, this is when $V=cst$. The correct wave-function is somewhat more complicated to compute and requires an analysis of the boundary conditions at $x=0,D$

Another way to see why your Hamiltonian does not commute with the momentum operator is that the potential $V$ and thus the Hamiltonian you have is not translation invariant, there are three distinct regions:

$$ 1.\quad x<0 \\ 2.\quad 0<=x<=D \\ 3.\quad D<x $$ and the potential is $$ V = \Theta(x)-\Theta(x-D) $$ where $\Theta$ is the Heaviside step function.

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